Matrix/Equivalent/Eigenvalue/Fact

Let ${}M$ be an ${}n\times n$-matrix over a field ${}K$, and let ${}B$ denote an invertible ${}n\times n$-matrix. Let ${}a\in K$.

Then an ${}n$-tuple ${}{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}$ is an

eigenvector of ${}M$ for the eigenvalue ${}a$ if and only if

${}{\begin{pmatrix}x'_{1}\\\vdots \\x'_{n}\end{pmatrix}}=B{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\,$

is an eigenvector of the matrix ${}BMB^{-1}$ for the eigenvalue ${}a$. In particular, ${}M$ and ${}BMB^{-1}$

have the same eigenvalues.

Proof, Write another proof